Answer
$$\eqalign{
& \left( {\text{a}} \right)x = 5 \cr
& \left( {\text{b}} \right){\text{Decreasing on: }}\left( { - \infty ,\infty } \right) \cr
& \left( {\text{c}} \right){\text{No relative extrema}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{x}{{x - 5}} \cr
& {\text{We have }}f\left( x \right) = \frac{x}{{x - 5}} \cr
& \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{x}{{x - 5}}} \right] \cr
& f'\left( x \right) = \frac{{\left( {x - 5} \right)\left( 1 \right) - x\left( 1 \right)}}{{{{\left( {x - 5} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{x - 5 - x}}{{{{\left( {x - 5} \right)}^2}}} \cr
& f'\left( x \right) = - \frac{5}{{{{\left( {x - 5} \right)}^2}}} \cr
& {\text{There are no points at which }}f'\left( x \right) = 0 \cr
& {\text{The derivative is not defined at }}x - 5 = 0,{\text{ so we obtain the critical }} \cr
& {\text{point }}x = 5 \cr
& \cr
& \left( {\text{b}} \right) \cr
& {\text{Set the intervals }}\left( { - \infty ,5} \right),\left( {5,\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$
\[\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty ,5} \right)}&{\left( {5,\infty } \right)} \\
{{\text{Test Value}}}&{x = 4}&{x = 6} \\
{{\text{Sign of }}f'\left( x \right)}&{{\text{ }}f'\left( 0 \right) < 0}&{{\text{ }}f'\left( 6 \right) = \frac{1}{3} < 0} \\
{{\text{Conclusion}}}&{{\text{Decreasing}}}&{{\text{Decreasing}}}
\end{array}\]
$$\eqalign{
& \left( {\text{c}} \right) \cr
& {\text{No relative minimum or relative maximum}} \cr} $$
